Problem: The equation of a circle $C$ is $x^2+y^2+14x-16y+77 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2+14x) + (y^2-16y) = -77$ $(x^2+14x+49) + (y^2-16y+64) = -77 + 49 + 64$ $(x+7)^{2} + (y-8)^{2} = 36 = 6^2$ Thus, $(h, k) = (-7, 8)$ and $r = 6$.